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Proof step 4 from proof gcdhex (previous step)

j = 0, j ≤ i, 0 ≤ j, 0 ≤ i, ∀ i0, j. i→n(i0) < i→n(i) → 0 ≤ i0 ∧ 0 ≤ j ∧ j ≤ i0 → (∃ i1, j0. gcd(i0, j) = i0 * i1 + j * j0) ⊦ ∃ i0, j0. gcd(i, j) = i * i0 + j * j0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: gcd-zero-04: ⊦ gcd(i, 0) = abs(i);%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%zero-left: ⊦ 0 * i = 0; from specification int-mult%absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs